The solubility product of Mg(OH)_2 is 1 times | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Rearrange the above expression to solve for $[OH^-]$: $[OH^{-}] = \sqrt{\frac

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Rearrange the above expression to solve for $[OH^-]$: $[OH^{-}] = \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}$ Substitute the given values ($K_{sp} = 1 	imes 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M$): $[OH^{-}] = \sqrt{\frac{1 	imes 10^{-12}}{0.01}} = \sqrt{1 	imes 10^{-10}} = 1 	imes 10^{-5} \ M$ Calculate $pOH$ from the hydroxide ion concentration: $pOH = -\log [OH^{-}] = -\log (1 	imes 10^{-5}) = 5$ Calculate $pH$ from $pOH$: $pH = 14 - pOH = 14 - 5 = 9$ Thus,precipitation will occur at $pH = 9$.

The solubility product of $Mg(OH)_2$ is $1 \times 10^{-12}$. At what $pH$ will precipitation occur in a $0.01 \ M$ $MgCl_2$ solution?

Similar Questions

If the concentration of $CrO_4^{2-}$ ions in a saturated solution of silver chromate is $2 \times 10^{-4} \ M$,the solubility product of silver chromate will be:

$A$ precipitate of $CaF_2$ $(K_{sp} = 1.7 \times 10^{-10})$ will be obtained when equal volumes of the following are mixed:

Up to what $pH$ must a solution containing a precipitate of $Cr(OH)_3$ be adjusted so that the precipitate just dissolves? (Given: $[Cr^{3+}] = 0.1 \ mol/L$,$K_{sp} = 6 \times 10^{-31}$)

Which of the following sulphides has the lowest solubility product?

The molar solubility of $CaF_2$ $(K_{sp} = 5.3 \times 10^{-11})$ in $0.1 \ M$ solution of $NaF$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module